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184_notes:examples:week7_wire_dimensions [2017/10/04 13:14] – [Example: Changing the Dimensions of a Wire] tallpaul | 184_notes:examples:week7_wire_dimensions [2017/10/04 13:51] – [Solution] tallpaul | ||
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====Solution==== | ====Solution==== | ||
- | Let's start with node $A$. Incoming current | + | Let's start with segment 1. The electric field is constant since the wire is uniform with respect to the rest of the segment, so we get $$E_1 = \frac{\Delta V_1}{L_1} = 0.5 \text{ |
- | $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 | + | The power dissipated through |
- | We do a similar analysis | + | Now, for segment 2. We can use [[184_notes: |
- | $$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$$ | + | |
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- | For node $C$, incoming current is $I_2$ and $I_3$. There is no outgoing current defined yet! $I_{C\rightarrow D}$ must be outgoing to balance. To satisfy the Node Rule, we set | + | |
- | $$I_{C\rightarrow D} = I_{out} | + | |
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- | Lastly, we look at node $D$. Incoming current is $I_{B\rightarrow D}$ and $I_{C\rightarrow D}$. Since there is no outgoing current defined yet, $I_{D\rightarrow battery}$ must be outgoing to balance. To satisfy the Node Rule, we set | + | |
- | $$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ | + | |
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- | Notice that $I_{D\rightarrow battery}=I_1$. This will always be the case for currents going in and out of the battery (approximating a few things that are usually safe to approximate, | + | |
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- | {{ 184_notes: | + |